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re: Help me freak out me daughter
Posted on 12/4/17 at 8:33 pm to PhillipJFry
Posted on 12/4/17 at 8:33 pm to PhillipJFry
quote:
Xy - 3X + 12 - 4y
X - 3X + 12 - 4
X - 3X + 8
-2X + 8
X = -4
0
Posted on 12/4/17 at 8:34 pm to ThatMakesSense
quote:
I'd enjoy watching you verbalize this to me.
'That 'y'..frick that'
I expect it would go something like this, particularly when Billy Blaze scratches out the first syllable. Night Shift- Prostitution scene
Posted on 12/4/17 at 8:41 pm to tigerdup07
For starters if there are two variables you need two equations.
Even assuming you set that expression equal to zero you still can’t solve it.
Even assuming you set that expression equal to zero you still can’t solve it.
Posted on 12/4/17 at 8:50 pm to tigerdup07
288
Posted on 12/4/17 at 9:08 pm to tigerdup07
X(y-3) + 4(3-y)
X*(y-3) + 4*(3-y) = 0
X*(y-3) = -4*(-1 *(3-y))
X*(y-3) = -4*(y-3)
X*(y-3) / (y-3) = -4*(y-3) / (y-3)
X = -4
y = 3
X*(y-3) + 4*(3-y) = 0
X*(y-3) = -4*(-1 *(3-y))
X*(y-3) = -4*(y-3)
X*(y-3) / (y-3) = -4*(y-3) / (y-3)
X = -4
y = 3
This post was edited on 12/4/17 at 9:10 pm
Posted on 12/4/17 at 9:09 pm to 3deadtrolls
quote:
xy-3x-4y+12
Damn i remember this shite.
Like rewriting the equation. I forget the official term for it
Posted on 12/4/17 at 9:19 pm to BayouBengals03
Mathing
Posted on 12/4/17 at 9:20 pm to CptRusty
quote:
For starters if there are two variables you need two equations.
Even assuming you set that expression equal to zero you still can’t solve it.
Actually if the equation is set to zero, it means either X or Y has to set the equation equal to zero in its respective factor, for example, in this equation which can be factored to (x-4)(y-3), if set equal to zero, either Y has to be equal to 3 or X has to be equal to 4 to set that factor to zero, which thereby multiplies the equation to zero giving the answer of zero. So in theory you can't solve the equation fully but you could give parameters to the solution.
Posted on 12/4/17 at 9:28 pm to RickAstley
quote:
X(y-3) + 4(3-y)
X*(y-3) + 4*(3-y) = 0
X*(y-3) = -4*(-1 *(3-y))
X*(y-3) = -4*(y-3)
X*(y-3) / (y-3) = -4*(y-3) / (y-3)
X = -4
y = 3
Wouldn't the right side of the equation be:
=-1(4(3-y))
=-1(12-4y)
=-12+4y
=-4(3+y)
In either case, there is more than one admissible solution with at least 2. There may be more, but here's 2 of them:
#1: x=-4; y=3
#2: x=4; y=0
This post was edited on 12/4/17 at 9:30 pm
Posted on 12/4/17 at 9:34 pm to Spock's Eyebrow
I know my equation is wrong. Just solved it incorrectly to get a number.
Posted on 12/4/17 at 9:36 pm to ThatMakesSense
quote:
Help me freak out me daughter by ThatMakesSense
Leave a floater in her toilet tomorrow.
Posted on 12/4/17 at 9:44 pm to AFtigerFan
quote:
quote:
X(y-3) + 4(3-y)
X*(y-3) + 4*(3-y) = 0
X*(y-3) = -4*(-1 *(3-y))
X*(y-3) = -4*(y-3)
X*(y-3) / (y-3) = -4*(y-3) / (y-3)
X = -4
y = 3
Wouldn't the right side of the equation be:
=-1(4(3-y))
=-1(12-4y)
=-12+4y
=-4(3+y)
In either case, there is more than one admissible solution with at least 2. There may be more, but here's 2 of them:
#1: x=-4; y=3
#2: x=4; y=0
Because you are dividing by a variable it is omitting infinitely many solutions. if set equal to zero anything along the graph of x=4; y=3 is correct.
This post was edited on 12/4/17 at 9:46 pm
Posted on 12/4/17 at 9:48 pm to tonydtigr
quote:
Because you are dividing by a variable it is omitting infinitley many solutions. if set equal to zero anything along the graph of x=4; y=3 is correct.
Good call, I didn't even catch the infinite amount of solutions. I guess more the more correct (or accurate?) thing to say would be anything along the graph of y=3 is correct, among others such as x=4; y=0.
This post was edited on 12/4/17 at 9:50 pm
Posted on 12/4/17 at 9:50 pm to tigerdup07
Ask Alexa.
Posted on 12/4/17 at 9:50 pm to tigerdup07
There’s a scary amount of folks in this thread without a basic understanding of the principles of math
Posted on 12/4/17 at 10:15 pm to tigerdup07
Xy = 8675309
If you daughter's name is Jenny.
If you daughter's name is Jenny.
Posted on 12/4/17 at 10:17 pm to tigerdup07
xy-3x + 12-4y. It’s called factoring by grouping
x(y-3) + -4(-3+y)
(x-4)(y-3)
x=4 , y=3
Also you could have just typed this in to mathway.com and done factor and it would have solved it
x(y-3) + -4(-3+y)
(x-4)(y-3)
x=4 , y=3
Also you could have just typed this in to mathway.com and done factor and it would have solved it
This post was edited on 12/4/17 at 10:31 pm
Posted on 12/4/17 at 11:50 pm to tigerdup07
Send me her snapchat, I'll traumatize the frick out of her.
Eta
Sorry, didn't read OP.
Eta
Sorry, didn't read OP.
This post was edited on 12/4/17 at 11:51 pm
Posted on 12/5/17 at 12:02 am to FloridaMike
quote:
There’s a scary amount of folks in this thread without a basic understanding of the principles of math
The only thing I have ever used math for is to count dollar bills. Other than that math is absolutely worthless.
This post was edited on 12/5/17 at 12:05 am
Posted on 12/5/17 at 12:09 am to MintBerry Crunch
quote:
Xy-4y-3x=-12
is she a fricking heathen?
multiply both sides by -1 jesus christ
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